Eric Nicholas K's answer to Candice's Secondary 4 A Maths Singapore question.
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Good evening Candice! Here are my workings for this question. We assume that the balls are drawn without replacement.
Date Posted:
4 years ago
Alternative approaches :
P(None of the balls are red)
= 1 - P(at least 1 red)
= 1 - P(1 red) - P(2 red) - P(3 red)
= 1 - (4/11 x 7/10 x 6/9 x 3) - (4/11 x 3/10 x 7/9 x 3) - (4/11 x 3/10 x 2/9)
= 1 - 28/55 - 14/55 - 4/165
= 1 - 26/33
= 7/33
b)
answer already obtained in a)
P(None of the balls are red)
= 1 - P(at least 1 red)
= 1 - P(1 red) - P(2 red) - P(3 red)
= 1 - (4/11 x 7/10 x 6/9 x 3) - (4/11 x 3/10 x 7/9 x 3) - (4/11 x 3/10 x 2/9)
= 1 - 28/55 - 14/55 - 4/165
= 1 - 26/33
= 7/33
b)
answer already obtained in a)
Thank you i understand now!
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